SE 5 Acids Bases pH and Molarity 19/2/18
Acids, Bases pH and
Molarity
Acids vs Bases
Acids Bases
• Release H+ ions in solution • Release OH- ions in solution
• Start with H and end with • Start with metal and end with
non-metal (except acetic and OH (except ammonium)
citric acid)
A. Di Lallo CRHS 2018 1
SE 5 Acids Bases pH and Molarity 19/2/18
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pH Scale
• Measure of the molar concentration of H+ ions in solution
• pH scale→ concentration of H+ ions written in scientific
notation
• Reason why each unit differs by a factor of 10
Concentration of H+ ions Concentration in scientific pH Handout
(mol/L) notation
1.0 1.0 × 100 0
0.1 1.0 × 10−1 1
0.01 1.0 × 10−2 2
0.001 1.0 × 10−3 3
0.0001 1.0 × 10−4 4
0.00001 1.0 × 10−5 5
0.000001 1.0 × 10−6 6
0.0000001 1.0 × 10−7 7
0.00000001 1.0 × 10−8 8
0.000000001 1.0 × 10−9 9
0.0000000001 1.0 × 10−10 10
0.00000000001 1.0 × 10−11 11
0.000000000001 1.0 × 10−12 12
0.0000000000001 1.0 × 10−13 13
0.00000000000001 1.0 × 10−14 14
A. Di Lallo CRHS 2018 2
SE 5 Acids Bases pH and Molarity 19/2/18
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Acid Base Neutralization
Acid + Base → Salt + Water
• The H+ released from the acid combines with the OH- released
from the base to form water
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Acid base neutralization and stoichiometry
• Stoichiometry of an acid/base neutralization = titration
procedure!
• Used to analyze the initial concentration of an unknown acid/base, by
measuring the volume of used substances in the neutralization reaction
A. Di Lallo CRHS 2018 3
SE 5 Acids Bases pH and Molarity 19/2/18
35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 mol/L
HCl by titration to an equivalence point. What is the
concentration of NaOH?
𝐻𝐶𝑙 + 𝑁𝑎𝑂𝐻 → 𝐻2 𝑂 + 𝑁𝑎𝐶𝑙 Mol HCl
ratio 1 1
0.0998 𝑚𝑜𝑙 𝑥
mol = = 0.002515 𝑚𝑜𝑙
0.002515 𝑚𝑜𝑙 0.002515 𝑚𝑜𝑙 1𝐿 0.0252 𝐿
rxn
Conc NaOH
𝑥 0.002515 𝑚𝑜𝑙
= = 0.0706 𝑚𝑜𝑙/𝐿
1𝐿 0.03562 𝐿
128.3 mL of Mg(OH)2 is neutralized with 42.5 mL of 0.25 mol/L
HCl by titration to an equivalence point. What is the
concentration of Mg(OH)2
2𝐻𝐶𝑙 + 𝑀𝑔(𝑂𝐻)2 → 𝐻2 𝑂 + 𝑀𝑔𝐶𝑙2 Mol HCl
ratio 2 1
0.25 𝑚𝑜𝑙 𝑥
mol = = 0.01069 𝑚𝑜𝑙
0.01069 𝑚𝑜𝑙 0.00535 𝑚𝑜𝑙 1𝐿 0.04275 𝐿
rxn
Conc NaOH
𝑥 0.00535 𝑚𝑜𝑙
= = 0.042 𝑚𝑜𝑙/𝐿
1𝐿 0.1283 𝐿
A. Di Lallo CRHS 2018 4